the ring $\frac{A[X]}{(X-a_{1},...,X-a_{n})}$ is isomorphic to A Can you please help me to show that : $$\frac{A[X]}{(X-a_{1},...,X-a_{n})}\simeq A$$ in which A is a ring and $a_{1},...,a_{n}\in A$.
thanks for your help.
 A: This is not true, e.g.:
$$\frac{\mathbb{Z}[X]}{(X-1,X-2)}\simeq 0\not\simeq \mathbb{Z}$$
simply because $X\equiv 1$ in the quotient (therefore any polynomial is equivalent to the sum of its coefficients), and $1\equiv 2$ therefore all elements of the quotient are equivalent to each other.

Edit: to be more precise, since $X-1\equiv 0$ in the quotient, and since $X^n-1$ is a multiple of $X-1$ for all $n$ 

$$X^n-1=(X-1)(X^{n-1}+\ldots +1)$$

it follows that $X^n-1$ is also $0$, in other words $X^n\equiv 1$ for all $n$. Therefore any polynomial $a_nX^n+\ldots a_1X+a_0$ is equivalent to the integer $a_n+\ldots +a_0$.
Now since $X\equiv 2$ and $X\equiv 1$, it follows that $1\equiv 2$ (so that $0\equiv 1$ by subtracting $0$ on each side) and therefore all integers are in turn equivalent to $0$. 

What you probably mean is that 
$$\frac{A[X_1,\ldots, X_n]}{(X_1-a_{1},...,X_n-a_{n})}\simeq A$$
am I right?
A: You can reach the quotient ring in two steps: let $I=(X-a_1)$, $J=(X-a_1,X-a_2,\dots,X-a_n)$; then
$$
A[X]/J\cong (A[X]/I)\big/(J/I)
$$
Now $A[X]/I\cong A$, via the homomorphism $\varphi\colon A[X]\to A$ defined by $\varphi(a)=a$, for $a\in A$, and $\varphi(X)=a_1$, which has kernel $I$.
If we consider $K=\varphi(J)$, we get that
$$
A[X]/J\cong A/K
$$
which is generally not isomorphic to $A$. You can check that
$$
K=(a_1-a_2,\dots,a_1-a_n)
$$
